Playtime with the hyperbolic plane

2 Feb

Update: Thanks to Anschel for noting that I messed up the statement of the last exercise.  It’s fixed now.  Thanks to Justin for noting that I messed up a square root.  Pythagorean theorem is hard, yo.

About a year and a half ago I explained what hyperbolic space is, specifically by contrasting it with Euclidean space and spherical space.  We’ve also run into hyperbolic groups a few times, which are groups whose Cayley graphs are somehow like hyperbolic space.  More precisely, a group is hyperbolic if, whenever you have a Cayley graph of that group, triangles are \delta-thin, which means the third side of any triangle is contained in a \delta neighborhood of the other two sides.  It’s important that the same \delta works for every triangle in the space.


Here the bottom side is contained in a neighborhood of the other two sides, and the triangle looks like it belongs in Star Trek


Here each side is contained in a small neighborhood of the other two sides, and it seems like the triangle is curving inward

Note that triangles in Euclidean space are way totally far from being \delta-hyperbolic.  For any big number n, you can make a triangle so that the third side is not contained in an n-neighborhood of the other two sides: just take a 2n horizontal segment and a 2n vertical segment to make an isoceles right triangle.  If is bigger than 2, then the midpoint of the hypotenuse is farther than away from the other two sides.  As usual, this long paragraph could be better done in a picture.


Soooo not hyperbolic: you can make arbitrarily fat triangles in Euclidean space.  Also, the purple line should have \sqrt{2}}n as its length, not the square root of n.[/caption]  I thought today we could just play around with hyperbolicity.  I'm running a small reading group on geometric group theory with some grad students, and today we got sidetracked a few times by just basic thoughts about geodesics in the hyperbolic plane.  We all thought they were interesting, so here I am trying to share it with you!  There are lots of other definitions of hyperbolicity, but I like latex \delta-$thin triangles.  Oh I forgot to mention that a nneighborhood of a point/line/shape consists of all the points within n of that point/line/shape.  So, for instance, a 3-neighborhood of a point in Euclidean space is a circle.  But with a taxicab metric, that 3-neighborhood is a squarey circle.

[caption id="attachment_3081" align="alignnone" width="181"]threeball Purple points are all distance three or less from red point

Anyways, I just put in that definition because it’s the first thing you’ll hear or see in a colloquium talk that involves the word “hyperbolic.”  Let’s play with the upper half plane model of hyperbolic space!  Here’s a repeat picture from that October 2014 post (wow that’s when baby was born!  He’s walking around and getting into trouble now, btw.).


Straight lines are ones that go straight up to infinity, and segments of half-circles whose diameters lie on the bottom line

The graph paper lines in this picture are misleading; they contrast hyperbolic geodesics with Euclidean ones.  So the gray lines are Euclidean geodesics, and the colored ones are hyperbolic.  All geodesics in this model are either straight lines perpendicular to the horizontal axis, or semicircles perpendicular to the horizontal axis.  All of the horizontal axis and everything that the straight up and down geodesics end at (sort of like a horizontal axis infinitely far away) represent infinity.

I’ll write down the metric in case you were wondering, but we won’t need it for what we’ll be doing: ds^2=\frac{dx^2+dy^2}{y^2} [I took this formulation from wikipedia].  What this says is that the hyperbolic metric is a lot like the Euclidean one, except that the higher up you go on the y-axis, the less distance is covered (because of that 1/y factor).  More precisely, if you’re just looking at the straight line geodesics, the distance between two points at heights a<b is ln(\frac{b}{a}).


All the lines have the same length ln(2).  Blue: ln (8/4), green: ln (16/8)

The other fact we might want to know is that things that look like Euclidean dilations (stretching something like your pupil dilates from looking in a bright light to a dark room) are isometries in this model. You can see that in the picture above: the lines look like they’re stretching longer and longer in the Euclidean metric, but they’re actually all the same length.  Speaking of isometries, if you have any two geodesics (like a vertical line and a big old semi-circle somewhere else), you can find an isometry that sends one to the other.

First question: what do circles look like?  Whenever you have a metric space, it’s nice to know what neighborhoods look like, and the first thing you might want to consider are neighborhoods of points.  Turns out circles in this model look like circles in Euclidean space, but the centers aren’t where you think they are.  For instance, here’s a picture of circles with radius ln(2), which we saw in the straight lines above.



The center of each circle is at the top of its surprised mouth.  The next highest line segment shows that each vertical diameter is actually a diameter (twice the radius).

Notice that the centers of these circles hang a lot lower in the circle than they do in the Euclidean metric!  Isn’t playtime fun?!

Generally when I play with math I throw out a lot of garbage ideas, and then eventually one of them is somewhat right.  Other people apparently think for awhile before they put out an idea.  Anyways, here are some sketches of what I thought a 2-neighborhood of a vertical line might look like:


This is the most subtle joke I have ever put in this blog

Maybe you looked at these and were like “Yen that is nonsense what were you thinking?!”  Maybe you are my advisor or a practiced mathematician.  Let’s go through the nonsense-ness of each of these pictures:

The rightmost picture is a 2-neighborhood of the vertical line in Euclidean space.  We know hyperbolic space is pretty drastically different from Euclidean space, so we wouldn’t expect the neighborhoods to be so similar.  The middle and left pictures have similar shapes but different curviness, and yes we’d expect a hyperbolic neighborhood to look different so those are guesses based in some more intution.  However, let’s try to figure out the actual size of a neighborhood of a vertical line.  We can use our previous pictures, and switch to a ln(2) neighborhood.


Changed my mind this is the most subtle joke I’ve put in this blog please someone get it and appreciate it please please

Here I moved all our ln(2) circles so that their centers laid on the same line.  A neighborhood of a line is just the union of the neighborhoods of all of the points on that line, so if we just keep making ln(2) circles along the line we’ll end up with a neighborhood of the whole line.  So you can see that our actual neighborhood ended up being upside down from my middle picture above.  If this explanation didn’t make sense, here’s [half] a 2-neighborhood of a Euclidean line:


Note how the denser the circles, the closer their boundaries on the left get to becoming that straight line we see on the right.

Actually using Euclidean intuitions and then mixing them up a bit is a great way to play with the hyperbolic plane.  This next exercise was an actual exercise in the book but it is just so crazy I have to share it with you.  It’s just dramatically different from Euclidean space, just like the triangles were.

If you have a circle in the hyperbolic plane and project it to a geodesic segment that it doesn’t intersect (which means for any point on the circle, you find the closest point to it on the geodesic and draw a dot on the geodesic there), the projection is shorter than ln(\frac{\sqrt{2}+1}{\sqrt{2}-1}).  Here’s the picture in Euclidean space where this makes no sense:


Third place likes getting on the podium.  I meant, the vertical lines show the projections from the faces to the horizontal line, and you can see they can be as big as you want if you just make bigger and bigger circles.

And here’s a picture in hyperbolic space that might make you think this could possibly just maybe be true.  Any circle will eventually fit inside a big huge circle that looks like the blue one in the picture, so its projection would be shorter than the projection of the blue one.  That means you only have to worry about big huge circles in that particular position.  And by “big huge,” I mean “of (Euclidean) radius n“.


Remember, if we’re just looking at vertical lines, we know how to measure distance: it’s ln(\frac{a}{b}).  So if you can show that the small orange circle hits the vertical line at \sqrt{2}n-n and the big orange circle hits it at \sqrt{2}n+n, you’ll have proved the contraction property.  Try using Euclidean geometry, and think about how we did the triangles case.

That was fun for me I hope it was fun for you!


2 Responses to “Playtime with the hyperbolic plane”

  1. Anschel Schaffer-Cohen February 3, 2016 at 3:43 pm #

    Trying to understand the argument at the end: am I right that this only works if we have some isometries that can transform any line into one of these “straight to infinity” kinds? What are those isometries?

    Also, what if the geodesic passes through the circle?

    • yenergy February 3, 2016 at 3:55 pm #

      Yes, great, I flew the fact that you need to map your geodesic to a vertical line under the radar. I actually left out the complex plane entirely in this post, but this upper half plane model of hyperbolic space means upper half of the complex plane, not Euclidean, if you’re actually going to make calculations. Fact: whenever you have two geodesics in the hyperbolic plane, you can build an isometry from one to the other. General isometries of the plane are given by SL(2,R). If you have a matrix (a b \\ c d), then the action of that matrix sends the point z to (az+b)/(cz+d). So the set up of this problem implies that step 1 is find an isometry to map your geodesic of choice to the imaginary axis. Step 2 is find an N large enough so that the image under this isometry of your ball of choice fits inside it.

      Thanks for the disjointedness catch I missed that! In fact it’s definitely not true if the circle and geodesic are not disjoint (just make a bigger and bigger circle over the geodesic). I’ll fix that in the post.

      On Wed, Feb 3, 2016 at 3:43 PM, Baking and Math wrote:


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